What is the extraneous solution to these equations? $\dfrac{x^2 - 3x}{x - 4} = \dfrac{3x - 8}{x - 4}$
Solution: Multiply both sides by $x - 4$ $ \dfrac{x^2 - 3x}{x - 4} (x - 4) = \dfrac{3x - 8}{x - 4} (x - 4)$ $ x^2 - 3x = 3x - 8$ Subtract $3x - 8$ from both sides: $ x^2 - 3x - (3x - 8) = 3x - 8 - (3x - 8)$ $ x^2 - 3x - 3x + 8 = 0$ $ x^2 - 6x + 8 = 0$ Factor the expression: $ (x - 2)(x - 4) = 0$ Therefore $x = 2$ or $x = 4$ At $x = 4$ , the denominator of the original expression is 0. Since the expression is undefined at $x = 4$, it is an extraneous solution.